# The Naked Scientists Forum

### Author Topic: Work and Energy, and Special Relativistic Equations - Lessons  (Read 9127 times)

#### Mr. Scientist

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##### Work and Energy, and Special Relativistic Equations - Lessons
« on: 05/01/2009 18:44:48 »
First we will work on Work and Energy. Then we will move onto Special Relativity.

The reason i broke this up, was to distinguish what equations i have numerically worked with, and what equations i have not. These lessons will commence from hereon, in this thread only, but i will continue General Relativity in the previous thread.

Thanks

#### Mr. Scientist

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #1 on: 05/01/2009 19:31:09 »
Work and Energy, Installation 1 of 3

Energy

The entire universe is filled with energy. In fact, there is an infinite amount of energy in the vacuum for the vacuum itself to manifest into real fluctuations. And there is a lot of energy in a piece of matter, in fact, it takes a lot of energy to create the solid world of matter, so you can get a lot of energy out of the matter in question.

Take a dice. A dice has about 10^80 grams of energy! Now that is a lot energy. That is more or less the same amount of particles we can measure in the observable universe. This is why, when we are able to split a simple atom, we can get a massive amount of energy, and the equations describing this lies in Mc². The c factor here that is squared, stands for ‘’celeritas,’’ or also been known as ‘’sceleritas,’’ which in latin means sunlight. The c factor here then, is the speed of light, which is about 186,350 miles per hour. That means that light will travel 186,350 miles in just one second!! In one year, light will travel 65 billion billion miles! The c in the equation of E=Mc² is very important to the theory of relativity, as it acts as a conversion element, which transforms the right hand side into the massive amount of energy we gain from the left.

The energy of matter, determines its ability to move (1). In fact, we call this energy ‘’Kinetic Energy,’’ and this is the energy a system will require to move. Thus Kinetic Energy is then said to be the energy of a moving system.

Nothing, in physics, is said to ever be totally stationary. At the atomic and subatomic level, matter is always in motion, and would continue to be in motion even if made to be in circumstances where there is an infinitely cold temperature, so you cannot ever totally freeze the movement of quantum systems. This cold temperature is referred to as the zero-point field of the vacuum, and it is itself an infinite density of energy found in every corner of the universe!

From a hurricane hitting the cost of some poor suburb of America, to your own ability to run after the ice cream van for your favourite treat, these actions require the kinetic energy resident in the body, or the energy content, which is simply mass times the speed of light squared Mc².

Kinetic Energy

For a mathematical description of a uniformly moving object, all we require is the following derivation:

v+v’/2=v*

Δx=vt,

a=v’-v/t

Which is suffice information describing a uniform acceleration. In the appearance of an observable motion, we find the following relationships:

Δx=v*t

Δx=v*(v’-v/a)

Δx=(v+v’/2)(v’-v/a)

*Note, the symbol here Δ is the Greek letter Delta, and it means ''a change in.''

Δx=v'²-v²/2a

So this remains a description of, as said, the motion in terms of observable qualities, the initial, final velocities, ect ect. Now we find a relationship with the force in the following;

aΔx=v’²-v²/2

Now we must associate these subjects with mass, so we say now

MaΔx=Mv’²-Mv²/2

What we did, was that we used the idea that F=Ma, one of Newtons laws saying that the mass multiplied by acceleration allows us to have a force exerted on an object. Even though we can reduce the above equation by substitution of p=Mv, which states the momentum is equal to the mass times velocity, giving by substitution,

MaΔx=p’²/2 – p²/2

We want to really solve in the following way,

FΔx = Mv’²/2 – Mv²/2

So we can solve the quantity to allow us to see the numerical relationship of the kinetic energy used for motion,

½ Mv² or ½ p², if you wish. This is the kinetic energy of a moving object. So we now say

KE=1/2Mv²

Where KE is the kinetic energy.

(1)   – to carefully distinguish the science here from my own discovery, I found that force and energy are indeed related to half the energy of a system, identical to the prediction of kinetic energy. When someone asks why something moves, we find that F=dt, where d equals distance and t equals time. We find the force then is exerted on a system, and it allows a thing to move. When under close consideration of the kinetic energy of a moving object, E_k=1/2 mv, we see that the momentum of a thing is found to be equal to half of the energy of the system. Therefore, my own calculation of motion due to force and energy if perfectly describable, and ultimately proves these equations that have been in the quantum bible for many years… My equation therefore says, ½(Fvt)²= ½ E² which is also derivable as ½(Fvt)²= ½ E², and if one wants to see these derivations, I am happy to show them.
« Last Edit: 05/01/2009 19:36:09 by Mr. Scientist »

#### daveshorts

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #2 on: 06/01/2009 11:32:22 »
Urrr... Yes.  This is all fairly standard.

I think your derivation of kinetic energy is pretty much how it was invented in the first place, it all comes out of calculus.

½ mv² is not equal to ½ p²

½ mv² = ½ p²/m

To apply a force F at a certain velocity v for a certain time you will have moved  a distance vt

E = F.d
E = F.vt
Where E is an energy

which is the same equation as you quote as
½(Fvt)²= ½ E²

So a standard result, and newtonian physics is remarkably self consistant!

#### Mr. Scientist

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« Reply #3 on: 06/01/2009 12:27:51 »
But i found it no coincidence Fvt came out of the derivation of 1/2 of the energy of system. That was independant of concluding the kinetic energy at the time. And yes, thanks for that correctio involving the momentum.

#### Mr. Scientist

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« Reply #4 on: 06/01/2009 17:49:22 »
Potential Energy

So after analysing the Kinetic Energy of a moving body, we will now investigate the potential energy concept in matter.

Work is expressed as the ability to make something happen, so for a system to have a work, there must be some potential inherent in the system. This is potential energy.

If we place an apple on the top step of some ladders that ascend up the side of a wall, then there is actually a lot we can say about the potential energy inherent within the system. Before we do investigate this, we can assume the following mathematical relationship:

KE+PE=C

Where KE is for the kinetic energy, and PE if for potential energy, and C is some constant. This constant is defined as (Mgh) which is also a mathematical expression, which states that mass multiplied by gravity and multiplied by height, yields the kinetic energy added with the potential energy. So if C=Mgh, then we can rearrange the above equation:

KE+PE=Mgh

Going back to our analogy, if the apple is on top of the ladders, it can now be said to have the potential energy to fall. This ghost energy would be lost, if the apple was sitting on the ground. If the apple fell from the top of the ladders, potential energy is then converted into kinetic energy, so a change in Kinetic Energy would imply a reducing of potential energy, and we can state this as:

Δ KE=Δ –PE

If we say the apple was falling with a uniform motion, we can also state that it has fallen h/2, which means it is then h/2 units above the ground, where the (h) stands for height. So then we can accumulate, if we freeze it momentarily in time:

-h/2=v’² - v²/-2g

Since gh=v’²  and v=0, since it is frozen in time.

The (g) here is for gravity, and then we say that it is accelerating towards the earth due to gravity, so gravity causes acceleration of objects. So kinetic energy can be given the relationship of:

KE=1/2M(gh)

We learn this by substitution of the following relationships:

KE=mgh/2

PE=mg(h/2) so

KE +PE=mgh/2+mgh/2=mgh

Which takes us back to the constant we began by assuming.  To help you understand how to use the equation, suppose you had a mass of 1200kg falling through the air, with a velocity of 23.0 m/s then what would the value of Kinetic Energy be, if you used the equation

KE=1/2Mv²

#### Mr. Scientist

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« Reply #5 on: 06/01/2009 19:26:46 »
Conservation of Energy

The non-scientist will have heard of conservation, but what is the conservational law of physics? What is being conserved?

Energy pervades all physical systems, even the fabric of the universe is an energy blanket. This is why we say the vacuum is a physical sheet. But energy cannot be lost in this vacuum. Energy does not simply ‘’vanish’’, but only change form. If a bit of energy makes some solid matter, that matter must be just a differential form of conserved energy, so that when the matter is converted back to energy, you end up with the same energy you began. So energy changes, but it never vanishes.

Some countable quality of a system never changes its value, even if its form might change, so the energy the universe began with at some distant point in the history light cone, is the same energy the universe has today (1).

We therefore say, that the conservation of energy is a fundamental law which the universe must always obey. If it didn’t, physical theory would fail as some understandable quality.

The law of conservation, in courses taken at university and college level, take the mathematical form of:

T_e=KE+PE

Which says that the total energy T_e is yielded equivalently as the Kinetic Energy added with the Potential Energy, a form of the equation we saw in the second installation.

Now we will investigate impulse, which is defined as nothing but a change in momentum, which is itself simply the measure of a mass in motion. These terms and buzz-words may be hard to pick up, but read this over again if you find anything hard to digest. Studying Impulse will help us analyze a well-used topic in physics, which is the conservation of momentum.

Impulse

According to Classical Mechanics, (which by definition is a theory of physics which does not take into accordance the Uncertainty Principle (2)), Impulse is calculated as the integral of the force exerted on the system, the integration is with respect to the time:

J=∫ F dt

Where (J) is for impulse, sometimes denoted as (I), F is for force, and dt is the change of infinitesimal time units. Calculating this is relatively easy with a little understanding into calculus,

J=∫ dp/dt (dt)

J=∫ dp

Which then results in

J=Δp

Where impulse is equivalent to a change in the momentum (p). Assuming the mass remains constant, as in non-relativistic speeds, then this can be shown as:

J=FΔt=MΔv=Δp

Since

p=Mv

Back to the Conservation Laws

Now let us imagine you had two snooker balls, and you rolled one to the other ball, one can quickly surmise that they will collide. Traditionally, we call the first ball M_1 and the second simply M_2.

For two objects, just like two snooker balls to collide, their impulses interact; this means that their impulses apply  to each other when some physical contact is at work. During such a collision, their impulses are equal in magnitude, but almost certainly in opposite directions, so we can mathematically state this as –

F_1Δt=-F_2Δt

(Here force is also assumed unchanged)

Which by substitution gives

Δp_1=Δ-p_2

So that would allow us to conclude

M_1v’_1-M_1v_1=M_2v’_2-M_2v_2

And if we substitute the right hand side for a change in momentum, we now have

M_1v’_1-M_1v_1=M_2v_2-M_2v’_2

After some mathematical simplification, we find that

p_t=M_1v_1+M_2v_2=0

Where p_t is the total momentum, and these equations state the momenta of the first term is equal to the momenta of the second action, and these state that even the momentum of a body is conserved.

(1)   – In fact, how can the universe have a specific measurable energy? For this to be true, there would need to be someone outside the universe to measure it! This is physically  true comment, and scientifically correct. It may change ones idea’s on how the universe manifests itself. Today, as there was just a few moments after the big bang (which was neither big, or a bang), there is about 10^80 particles, give or take a few tens. These particles are trapped forms of energy*, and these forms of energy has the exact same value they had before their transformations.

•   In fact there is a theory which states that matter is but differential forms of trapped light. The idea is that even though we have many forms of different energy, such as gluon energy, the kind of subatomic energy that binds all of matter, solid matter which can be converted back to energy through antimatter relationships reducing it back to photon energy. So the Luxon Theory of matter (first proposed by Newton), states that high energy conditions (not so much collisions), created all the forms of solid and tangible matter we observe today.

(2)   – The Uncertainty Principle was created by Mathematician and Physicist, Heisenberg in 1926, and his principle states that it is physically impossible for some observer to ultimately define complimentary states simultaneously, such as the position and the momentum of a physical system. His law is given as Δx•Δp ≥ h/2, where x is the position, and p is the momentum of the body. Define x with great accuracy, and then p is reduced as even more undefined. In fact, if you could the position x with total certainty, then the system will be made to have an infinite amount of momentum. Likewise, define p with certainty, then it is made to have an infinite amount of positions! You could know for certainty these values, if you are willing to observe the infinite qualities of position of momentum, which for a human observer is impossible.
« Last Edit: 06/01/2009 23:29:58 by Mr. Scientist »

#### Mr. Scientist

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #6 on: 06/01/2009 19:27:30 »
And that completes the first lesson on Energy.

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #7 on: 07/01/2009 16:57:45 »
Quote
KE+PE=C

Where KE is for the kinetic energy, and PE if for potential energy, and C is some constant. This constant is defined as (Mgh) which is also a mathematical expression, which states that mass multiplied by gravity and multiplied by height, yields the kinetic energy added with the potential energy. So if C=Mgh, then we can rearrange the above equation:

KE+PE=Mgh

What?? mgh is a formula not a constant, and why do you say KE+PE=Mgh? It's just PE=mgh.

#### lightarrow

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #8 on: 07/01/2009 19:55:10 »
Quote
KE+PE=C

Where KE is for the kinetic energy, and PE if for potential energy, and C is some constant. This constant is defined as (Mgh) which is also a mathematical expression, which states that mass multiplied by gravity and multiplied by height, yields the kinetic energy added with the potential energy. So if C=Mgh, then we can rearrange the above equation:

KE+PE=Mgh
What?? mgh is a formula not a constant, and why do you say KE+PE=Mgh? It's just PE=mgh.
There are many others imprecisions in his lesson. It would need a lot of time to correct all of it (and so it's probable there are in the others lessons).

#### lyner

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« Reply #9 on: 07/01/2009 23:58:37 »
Why is he posting this stuff?
It is all textbook material (where it's correct) and it wasn't in response to any particular question.
I think it should be on his own website then he wouldn't get any flack.

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #10 on: 08/01/2009 06:26:33 »
At least its held to criticism here so that those who don't know better are made aware of its inaccuracies.

#### Mr. Scientist

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« Reply #11 on: 08/01/2009 16:02:40 »
(Well actually, it may be wise to some posters to just raise their voices on me explaining the parts they don't understand, so that perhaps i could simply answer them. As i will do shortly.)

Sophie, that's all very true, but some people believe it or not out there cannot afford an education, never mind affording these textbooks you mention.

#### Mr. Scientist

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« Reply #12 on: 08/01/2009 16:14:12 »
It is certainly reasonable, according to my lecturer to assume that under strict conditions one can say that by adding the kinetic energy with the potential will yield some constant. In this case i have given, the constant value is given (mgh), and the conclusion is sound. To prove this is simple:

If KE=1/2M(4/3gh)=2/3mgh so PE=mg(h/3) then PE+KE=mgh.

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #13 on: 08/01/2009 17:23:07 »
If an object of 5kg is 30m high with gravity of 9.8^2 it will have the potential energy of 14406 joules of energy, using the formula PE=mgh.

If the same object was also travelling sideways at say 6 metres per second (KE=0.5*5*6^2) it will have a KE of 90 joules.

So PE + KE (14406 + 90) is 14496 joules.

so if PE + KE = mgh, then

14496 = 5 * 9.8^2 * 30
therefore 14496 = 14406???

How does that work?

#### lyner

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« Reply #14 on: 08/01/2009 17:52:55 »
Quote
so if PE + KE = mgh, then

But it doesn't; it equals the total energy of the object: =PE + KE .
If / when it hits the ground, its PE will be zero and its KE will have increased by the amount of PE it started with.
KE(final) = PE(initial) + KE(initial)

For an object in orbit (elliptical, in the general case), the total energy remains the same. It goes faster when it is nearer and slower when further away.
« Last Edit: 08/01/2009 18:12:19 by sophiecentaur »

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« Reply #15 on: 08/01/2009 18:31:00 »
Yeah I know it doesn't, that's what I'm saying

To figure out just the potential energy the formula is PE=mgh

#### Mr. Scientist

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #16 on: 08/01/2009 19:20:18 »
Quote
so if PE + KE = mgh, then

But it doesn't; it equals the total energy of the object: =PE + KE .
If / when it hits the ground, its PE will be zero and its KE will have increased by the amount of PE it started with.
KE(final) = PE(initial) + KE(initial)

For an object in orbit (elliptical, in the general case), the total energy remains the same. It goes faster when it is nearer and slower when further away.

The process or mathematical notation being used including the conclusions are not unheard of, well, at least in my countries educationaln system:

If you take into consideration the facts i gave you:

If we say the apple was falling with a uniform motion, we can also state that it has fallen h/2, which means it is then h/2 units above the ground, where the (h) stands for height. So then we can accumulate, if we freeze it momentarily in time:

-h/2=v’² - v²/-2g

(now)

Assume this equation is just so, because the object has fallen due to the acceleration of gravity, where 4/3gh=v'² where v=0.

This educational teaching is thus used in a book by Johnnie T Denise and Gary Moring, ''the complete guide to physics,'' so here is my proof you have either picked me up wrong, or you don't realise you are wrong about something.

#### lyner

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« Reply #17 on: 08/01/2009 21:39:30 »
I was using the usual notation for what's involved. You can make up your own system if you like but I can't understand what you're saying if you don't define the terms you use.

If these are supposed to be 'Science lessons' then I suggest you tidy your ideas up a bit before you try to teach others.

Quote
-h/2=v’² - v²/-2g

would seem to be dimensionally inconsistent, for a start. Perhaps there's something you don't understand or you left out some brackets and a factor of 2?
« Last Edit: 08/01/2009 21:42:55 by sophiecentaur »

#### Mr. Scientist

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« Reply #18 on: 08/01/2009 22:06:56 »
I am looking at a book which uses the same mathematical notation as i was once taught. The dimensions are not wrong, because -h/2 is the ''maxiumum'' height, so maybe you have not understood the math?

#### lyner

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« Reply #19 on: 08/01/2009 22:21:40 »
If you put brackets (parentheses) around the vsquared terms then, of course, the equation is correct. But I can't see this is the slightest bit ground breaking.
What actual point are you trying to make? Is there something new?
The 4/3 bit seems to come from nowhere, though.

btw, I presume that you really meant to say that the object is falling with uniform acceleration not uniform motion.
You really must try to be more accurate in what you write.
« Last Edit: 08/01/2009 22:27:50 by sophiecentaur »

#### Mr. Scientist

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« Reply #20 on: 08/01/2009 22:29:21 »
Are you just intentionally being a @bitch?

Trust me, most mathematical textbooks you will read that help people learn physics will not be groundbreaking, so once again, in find someone picking at things that are attemtping to degrade me in some way.

Give me a break, before i give you one.

#### lyner

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #21 on: 08/01/2009 22:36:40 »
Most textbooks are checked and proof read before they are published.
If they are not correct then they can't help anyone - in fact they can seriously damage the understanding of a student.

And there is no need to be rude.

#### Mr. Scientist

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« Reply #22 on: 08/01/2009 22:42:51 »
Yes?

Take a page out of your own book then, and do not be rude. Consequentially, i can assure you it is right, because you see it is right, so welcome to the circular logic you created.

#### lyner

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« Reply #23 on: 08/01/2009 22:54:04 »
Could you tell me how the 4/3 bit is 'right' or relevant?

#### Mr. Scientist

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« Reply #24 on: 08/01/2009 22:58:47 »
It is a changing length from the ground where v=0. I give up.

Simply this is the way we where shown to calculate a distance from the ground.

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##### Work and Energy, and Special Relativistic Equations - Lessons
« Reply #24 on: 08/01/2009 22:58:47 »